Control System Analysis Techniques course error

In the Time domain section of the course when I use the step function to plot the grpah it keeps saying i am wrong even though it is the exact same code and graph. Those anyone have anny solutions

below is the code

s = tf('s');
quadcopter_ol = 0.04133/(s + 0.01479);
controller = zpk(-5,0,100);

Task 1

step(quadcopter_ol)

[Walter Roberson]

Is it look for step(controller) ?

s = tf('s');
quadcopter_ol = 0.04133/(s + 0.01479);
controller = zpk(-5,0,100);
step(controller)

[Dunbarin]

No it’s you’re meant to use the step function on the quadcopter, even the solution in the course does the same thing I do and the graphs are the same but it keeps saying I’m wrong

Answer 1

[Dunbarin]

I reached out to technical support and they were able to give me a work around to the problem,I will paste it below

Use stepplot function instead of step function as follows:

output = stepplot(<variable_name>); output.Responses.Name = '<variable_name>';

Similarly, the following function mapping + output.Responses.Name workaround can be used to get around the issue in other parts of the course:

bode -> bodeplot nyquist -> nyquistplot pzmap -> pzplot

[Sam Chak]

Hi @Dunbarin

Thanks for the update. Virtually no difference between "step(sys)" and "stepplot(sys)", but the grading system can only accept one answer. The "stepplot(sys, plotoptions)" command can be used if targeted customization of the step plot is required, but "step(sys, timeDuration)" is generally good enough for initial visualization of the step response.

s = tf('s');
% Open-loop system
quadcopter_ol = 0.04133/(s + 0.01479)

quadcopter_ol =
 
    0.04133
  -----------
  s + 0.01479
 
Continuous-time transfer function.

% method 1
figure(1)
step(quadcopter_ol)

% method 2
figure(2)
stepplot(quadcopter_ol)

[Agasthya]

It seems to work for me. Thankyou

Answer 2

[Sam Chak]

Hi @Dunbarin

If this topic is related to control systems, you are probably required to implement the controller on the open-loop quadcopter to form a closed-loop control system with the desired characteristics, such as the settling time.

s = tf('s');

% Open-loop system
quadcopter_ol = 0.04133/(s + 0.01479)

quadcopter_ol =
 
    0.04133
  -----------
  s + 0.01479
 
Continuous-time transfer function.

% Controller
controller = zpk(-5,0,100)

controller =
 
  100 (s+5)
  ---------
      s
 
Continuous-time zero/pole/gain model.

% Closed-loop system
closedLoop = feedback(controller*quadcopter_ol, 1)

closedLoop =
 
       4.133 (s+5)
  ----------------------
  (s^2 + 4.148s + 20.66)
 
Continuous-time zero/pole/gain model.

% Step response of the closed-loop system
step(closedLoop), grid on

Warning: Graphics acceleration hardware is unavailable. Graphics quality and performance might be diminished. See MATLAB System Requirements.

[Dunbarin]

Thank you for your reply but the task specifically askes you to use the step function on quadcopter_ol, and what you did is the immediate task after completing what I said.

My issue is that I used the step function correctly but it still says I failed

[Sam Chak]

Hi @Dunbarin

I’m sorry to hear that. Have you resolved the issue? If not, please contact technical support for assistance.

[Dunbarin]

No I haven't resolved the issue yet but I have reached to tech support

[Andre]

Is there anything yet on this issue? I'm stuck on it.

[Elijah]

I am trying to do the same but it isn't working and I am stuck on it.